# All vertex pairs connected with exactly k edges in a graph

Given a directed graph represented as an adjacency matrix and an integer ‘k’, the task is to find all the vertex pairs that are connected with exactly ‘k’ edges.

Also, find the number of ways in which the two vertices can be linked in exactly k edges.

**Examples :**

Input :k = 3 and graph : 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 0 0Output :1 -> 4 in 1 way(s) 1 -> 5 in 1 way(s) 2 -> 1 in 1 way(s) 2 -> 3 in 1 way(s) 3 -> 2 in 1 way(s) 3 -> 4 in 1 way(s) 3 -> 5 in 1 way(s) 4 -> 3 in 1 way(s) 5 -> 1 in 1 way(s) 5 -> 3 in 1 way(s)Input :k = 2 and graph : 0 0 0 1 0 1 0 1 0Output :2 -> 2 in 1 way(s) 3 -> 1 in 1 way(s) 3 -> 3 in 1 way(s)

**Approach :**

- We will multiply the adjacency matrix with itself ‘k’ number of times.
- In the resultant matrix,
`res[i][j]`

will be the number of ways in which vertex ‘j’ can be reached from vertex ‘i’ covering exactly ‘k’ edges.

Below is the implementation of the above approach :

`// Java implementation of the approach ` `public` `class` `KPaths { ` ` ` ` ` `// Function to multiply two square matrices ` ` ` `static` `int` `[][] multiplyMatrices(` `int` `[][] arr1, ` `int` `[][] arr2) ` ` ` `{ ` ` ` `int` `order = arr1.length; ` ` ` `int` `[][] ans = ` `new` `int` `[order][order]; ` ` ` `for` `(` `int` `i = ` `0` `; i < order; i++) { ` ` ` `for` `(` `int` `j = ` `0` `; j < order; j++) { ` ` ` `for` `(` `int` `k = ` `0` `; k < order; k++) { ` ` ` `ans[i][j] += arr1[i][k] * arr2[k][j]; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` `return` `ans; ` ` ` `} ` ` ` ` ` `// Function to find all the pairs that ` ` ` `// can be connected with exactly 'k' edges ` ` ` `static` `void` `solve(` `int` `[][] arr, ` `int` `k) ` ` ` `{ ` ` ` `int` `[][] res = ` `new` `int` `[arr.length][arr[` `0` `].length]; ` ` ` ` ` `// copying arr to res, ` ` ` `// which is the result for k=1 ` ` ` `for` `(` `int` `i = ` `0` `; i < res.length; i++) ` ` ` `for` `(` `int` `j = ` `0` `; j < res.length; j++) ` ` ` `res[i][j] = arr[i][j]; ` ` ` ` ` `// multiplying arr with itself ` ` ` `// the required number of times ` ` ` `for` `(` `int` `i = ` `2` `; i <= k; i++) ` ` ` `res = multiplyMatrices(res, arr); ` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < res.length; i++) ` ` ` `for` `(` `int` `j = ` `0` `; j < res.length; j++) ` ` ` ` ` `// if there is a path between 'i' ` ` ` `// and 'j' in exactly 'k' edges ` ` ` `if` `(res[i][j] > ` `0` `) ` ` ` `System.out.println(i + ` `" -> "` `+ j + ` `" in "` `+ res[i][j] + ` `" way(s)"` `); ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `[][] arr = ` `new` `int` `[` `5` `][` `5` `]; ` ` ` `arr[` `0` `][` `1` `] = ` `1` `; ` ` ` `arr[` `1` `][` `2` `] = ` `1` `; ` ` ` `arr[` `2` `][` `3` `] = ` `1` `; ` ` ` `arr[` `2` `][` `4` `] = ` `1` `; ` ` ` `arr[` `3` `][` `0` `] = ` `1` `; ` ` ` `arr[` `4` `][` `2` `] = ` `1` `; ` ` ` `int` `k = ` `3` `; ` ` ` `solve(arr, k); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

**Output:**

0 -> 3 in 1 way(s) 0 -> 4 in 1 way(s) 1 -> 0 in 1 way(s) 1 -> 2 in 1 way(s) 2 -> 1 in 1 way(s) 2 -> 3 in 1 way(s) 2 -> 4 in 1 way(s) 3 -> 2 in 1 way(s) 4 -> 0 in 1 way(s) 4 -> 2 in 1 way(s)

The time complexity of the above code can be reduced for large values of k by using matrix exponentitation. The complexity can be changed from **O(n^3 * k)** to **O(n^3 * log k)**

`class` `KPaths { ` ` ` ` ` `// Function to multiply two square matrices ` ` ` `static` `int` `[][] multiplyMatrices(` `int` `[][] arr1, ` `int` `[][] arr2) ` ` ` `{ ` ` ` `int` `order = arr1.length; ` ` ` `int` `[][] ans = ` `new` `int` `[order][order]; ` ` ` `for` `(` `int` `i = ` `0` `; i < order; i++) { ` ` ` `for` `(` `int` `j = ` `0` `; j < order; j++) { ` ` ` `for` `(` `int` `k = ` `0` `; k < order; k++) { ` ` ` `ans[i][j] += arr1[i][k] * arr2[k][j]; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` `return` `ans; ` ` ` `} ` ` ` ` ` `// Function to find all the pairs that ` ` ` `// can be connected with exactly 'k' edges ` ` ` `static` `void` `solve(` `int` `[][] arr, ` `int` `k) ` ` ` `{ ` ` ` `int` `[][] res = ` `new` `int` `[arr.length][arr[` `0` `].length]; ` ` ` ` ` `res = power(arr, k, arr[` `0` `].length); ` ` ` `for` `(` `int` `i = ` `0` `; i < res.length; i++) ` ` ` `for` `(` `int` `j = ` `0` `; j < res.length; j++) ` ` ` ` ` `// if there is a path between 'i' ` ` ` `// and 'j' in exactly 'k' edges ` ` ` `if` `(res[i][j] > ` `0` `) ` ` ` `System.out.println(i + ` `" -> "` `+ j + ` `" in "` `+ res[i][j] + ` `" way(s)"` `); ` ` ` `} ` ` ` ` ` `static` `int` `[][] power(` `int` `x[][], ` `int` `y, ` `int` `n) ` ` ` `{ ` ` ` `// MATRIX EXPONENTIATION ` ` ` `// Initialize result ` ` ` `int` `res[][] = identity(n); ` ` ` ` ` `while` `(y > ` `0` `) { ` ` ` ` ` `if` `((y & ` `1` `) == ` `1` `) ` ` ` `res = multiplyMatrices(res, x); ` ` ` ` ` `// y must be even now ` ` ` `// y = y / 2 ` ` ` `y = y >> ` `1` `; ` ` ` `x = multiplyMatrices(x, x); ` ` ` `} ` ` ` `return` `res; ` ` ` `} ` ` ` `static` `int` `[][] identity(` `int` `n) ` ` ` `{ ` ` ` `// returns identity matrix of order n ` ` ` `int` `r[][] = ` `new` `int` `[n][n]; ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `r[i][i] = ` `1` `; ` ` ` ` ` `return` `r; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `[][] arr = ` `new` `int` `[` `5` `][` `5` `]; ` ` ` `arr[` `0` `][` `1` `] = ` `1` `; ` ` ` `arr[` `1` `][` `2` `] = ` `1` `; ` ` ` `arr[` `2` `][` `3` `] = ` `1` `; ` ` ` `arr[` `2` `][` `4` `] = ` `1` `; ` ` ` `arr[` `3` `][` `0` `] = ` `1` `; ` ` ` `arr[` `4` `][` `2` `] = ` `1` `; ` ` ` `int` `k = ` `3` `; ` ` ` `solve(arr, k); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

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